CONTROL SYSTEMS ENGINEERING D227 SOLUTIONS Q 2003

(a) Sketch the root locus of a unity feed back control system with the open loop transfer function

8)3)(ss(s

kG(ol)++

= For what value of k is the system stable?

(b) If a zero at s = -2 is introduced into the function of part (a), obtain the modified root locus. Describe the effect that the introduction of the zero has upon the performance. SOLUTION (a)

(a) 1G(ol)

G(ol)G(cl) 8)3)(ss(s

kG(ol)+

=++

=

The characteristic equation is 1 8)3)(ss(s

k+

++

There are three poles at 0, -3 and -8 For three poles the asymptotes form an angle of 60o and –60o to the real axis. The intercept with the real axis is at (0 – 3 – 8)/3 = 3.67 The value of k at the point where the locus cuts the imaginary axis is found from the Routh Hurwitz criteria. The characteristic equation may be written as:- s(s+3)(s+8) + k = 0 multiply out and s3 + 11s2 + 24s1 + kso

The highest power is 3 so use the simplified test. a = 1 b = 11 c = 24 d = k R = c - ad/b = 24 - k/11 and this is zero when the locus cuts the imaginary axis. 24 = k/11 k = 264 Any value of k larger than this produces instability. The asymptotes may now be drawn. One locus starts at -8 (k = 0) and moves to –∞ along the real axis. The other loci start at 0 and -3 and meet at some point before breaking away at 90o as mirror images and eventually forming asymptotes. It would be better if we knew the break away point. We are finding the roots of s3 + 11s2 + 24s1 + kso. If a calculator that solves roots is available then the roots can be solved for various values of k and a precise plot obtained as shown.

(b) With a zero at -2 the intercept of the asymptotes is {(0 – 3 – 8) – (-2)}/2 = 4.5 One loci starts at 0 and ends at -2 The other loci start at -8 and ends at - 4.5 + j∞ one and the other starts at -3 and ends at - 4.5 - j∞ The angle of the asymptotes is 90o to the real axis The system is stable for all values of k and less oscillatory as the gain is increased.